Past exam of the physics 2 paper of the University of Oxford / 2015 resit / 11 / a

For two identical narrow slits under coherent normal illumination, with $\lambda$ the wavelength in the propagation medium, let $I_s$ be the intensity contributed by either slit alone in the relevant angular range. At far-field angle $\theta$, the path difference is $\Delta =s\sin \theta$ and the phase difference is $2\pi \Delta /\lambda$. Adding the two complex amplitudes givesFor a screen perpendicular to the optical axis, $\tan \theta =z/d$. Thus the angular fringe factor uses $\sin \theta =z/\sqrt{d^2+z^2}$; in the paraxial region $|z|\ll d$ it reduces toThe supplied large-distance condition gives the Fraunhofer diffraction regime for the slit separation; the further replacement $\sin \theta \simeq z/d$ is a separate small-angle approximation. An unspecified finite slit width would add a single-slit envelope, and $I_s$ can then vary with angle. The formula above gives the interference factor in the ideal narrow-slit, equal-amplitude model; no absolute source intensity is supplied.